The post office does not sell any boxes which have a side length longer than 10 centimeters.
I bought a box from the post office that has a volume of VOL cubic centimeters.
So the dimensions of the package should be
FACTORIZATION[0],
FACTORIZATION[1] and
FACTORIZATION[2].
We can find the dimensions of the package by arranging the factors into three groups of products, all less than 10.
One solution would be:
(getPrimeFactorization(WIDTH).join("\\space\\color{black}{\\times}\\space")),
(getPrimeFactorization(LENGTH).join("\\space\\color{black}{\\times}\\space")) and
(getPrimeFactorization(HEIGHT).join("\\space\\color{black}{\\times}\\space"))
So the dimensions of the box could be WIDTH,
LENGTH and HEIGHT.
A water tank is a rectangular prism that is H1plural_form(METERS, H1) high,
L1plural_form(METERS, L1) long,
and W1plural_form(METERS, W1) wide.
A solid metal box which is H2plural_form(METERS, H2) high,
L2plural_form(METERS, L2) long,
and W2plural_form(METERS, W2) wide is sitting inside the tank.
The tank is filled with water.
The tank has a volume of H1 \times L1 \times W1 =
H1 * L1 * W1 cubic plural_form(METERS, H1 * L1 * W1),
and the metal box has a volume of H2 \times L2 \times W2 =
H2 * W2 * L2 cubic plural_form(METERS, H2 * L2 * W2).
Since there is no water in the box, the volume of the water in the tank is the volume of the tank minus the volume of the metal box.
The volume of the water in the tank is
H1 * L1 * W1 \text{UNIT}^3 -
H2 * L2 * W2 \text{UNIT}^3 =
H1 * L1 * W1 - H2 * L2 * W2\text{UNIT}^3
I made a fort for my pet gorilla by connecting two boxes.
The first box is
L1plural_form(METERS, L1) long,
W1plural_form(METERS, W1) wide and
H1plural_form(METERS, H1) high.
The second box is
L2plural_form(METERS, L2) long,
W2plural_form(METERS, W2) wide and
H2plural_form(METERS, H2) high.
How many cubic plural_form(METERS, 2) of space does my gorilla have to play in her fort?
The first box has a volume of H1 \times L1 \times W1 =
VOL1 cubic plural_form(METERS, VOL1).
The second box has a volume of H2 \times L2 \times W2 =
VOL2 cubic plural_form(METERS, VOL2)
Since my gorilla can play in both boxes, we need to add the volumes of the two boxes.
The total amount of space my gorilla has to play is
VOL1 \text{UNIT}^3 +
VOL2 \text{UNIT}^3 =
VOL1 + VOL2\text{UNIT}^3
How many cubic plural_form(TO) fit in VOL cubic plural_form(FROM, VOL)?
VOL * CONVERSION_CUBED cubic plural_form(TO)
First we will figure out how many cubic plural_form(TO) fit in one cubic FROM.
There are CONVERSIONplural_form(TO, CONVERSION) in every FROM.
So a cubic FROM is the same as a
CONVERSION \text{ TO_TEXT}
\times CONVERSION \text{ TO_TEXT}
\timesCONVERSION \text{ TO_TEXT}
cube.
init({range: [[-2, 2], [-2, 2]], scale: [100, 100]});
addMouseLayer();
//Adding verticies for the slabs
for (var i = 1; i < HEIGHT;i++ ){
VERTICES.push( [LENGTH/MAX,WIDTH/MAX,(-HEIGHT+2*i)/MAX]);
VERTICES.push([-LENGTH/MAX,WIDTH/MAX,(-HEIGHT+2*i)/MAX]);
VERTICES.push([-LENGTH/MAX,-WIDTH/MAX,(-HEIGHT+2*i)/MAX]);
VERTICES.push([LENGTH/MAX,-WIDTH/MAX,(-HEIGHT+2*i)/MAX]);
};
//Adding verticies for the cubes in one slab
for (var i = 1; i < LENGTH; i++ ){
VERTICES.push( [(-LENGTH+2*i)/MAX,-WIDTH/MAX,(-HEIGHT)/MAX]);
};
for (var i = 1; i < LENGTH; i++ ){
VERTICES.push( [(-LENGTH+2*i)/MAX,WIDTH/MAX,(-HEIGHT)/MAX]);
};
for (var i = 1; i < WIDTH; i++ ){
VERTICES.push( [(-LENGTH)/MAX,(-WIDTH+2*i)/MAX,(-HEIGHT)/MAX]);
};
for (var i = 1; i < WIDTH; i++ ){
VERTICES.push( [(LENGTH)/MAX,(-WIDTH+2*i)/MAX,(-HEIGHT)/MAX]);
};
for (var i = 1; i < LENGTH; i++ ){
VERTICES.push( [(-LENGTH+2*i)/MAX,-WIDTH/MAX,(-HEIGHT+2)/MAX]);
};
for (var i = 1; i < LENGTH; i++ ){
VERTICES.push( [(-LENGTH+2*i)/MAX,WIDTH/MAX,(-HEIGHT+2)/MAX]);
};
for (var i = 1; i < WIDTH; i++ ){
VERTICES.push( [(-LENGTH)/MAX,(-WIDTH+2*i)/MAX,(-HEIGHT+2)/MAX]);
};
for (var i = 1; i < WIDTH; i++ ){
VERTICES.push( [(LENGTH)/MAX,(-WIDTH+2*i)/MAX,(-HEIGHT+2)/MAX]);
};
// graphing the initial object
graph.obj = make3dObject(VERTICES, { scale: SCALE });
_.each(FACES, function(face) {
graph.obj.addFace(face);
});
graph.obj.setPos([0, 0, 5]);
graph.obj.rotate(1,0 , 0, PI/3);
graph.obj.rotate(0, 1, 0, 0);
graph.obj.rotate(0, 0, 1, PI/8);
graph.obj.doDraw();
var mouseTarget = mouselayer.rect(0, 0, 400, 400).attr({
fill: "#000", opacity: 0.0
});
$(mouseTarget[0]).bind("vmousedown", function(event) {
graph.lastX = event.pageX;
graph.lastY = event.pageY;
$(document).bind("vmousemove vmouseup", function(event) {
event.preventDefault();
var dx = (event.pageX - graph.lastX) / 200 * PI;
var dy = (event.pageY - graph.lastY) / 200 * PI;
graph.lastX = event.pageX;
graph.lastY = event.pageY;
var xAxis = normalize(graph.obj.perspective[1]);
var yAxis = normalize(graph.obj.perspective[0]);
graph.obj.rotate(xAxis[0], xAxis[1], xAxis[2], -dx / 2);
graph.obj.rotate(yAxis[0], yAxis[1], yAxis[2], -dy / 2);
graph.obj.doDraw();
if (event.type === "vmouseup") {
$(document).unbind("vmousemove vmouseup");
}
})
});
There are CONVERSION \times CONVERSION
\times CONVERSION = CONVERSION_CUBED \text{ TO_TEXT}^3
in each cubic FROM, by the volume formula.
Since we have VOL cubic plural_form(FROM, VOL),
and each cubic FROM holds CONVERSION_CUBED cubic
plural_form(TO, CONVERSION_CUBED), we have a total of
VOL \times CONVERSION_CUBED
cubic plural_form(TO, VOL * CONVERSION_CUBED).
So VOL * CONVERSION_CUBED cubic plural_form(TO, VOL * CONVERSION_CUBED)
fit in VOL cubic plural_form(FROM, VOL).